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Answer :
Final answer:
The resistance of the lightbulb connected to a 60 Hz power source and using an average power of 38.8 W, with a maximum voltage of 156 V, is approximately 312.4 Ω.
Explanation:
To find the resistance of the lightbulb, we must first establish the effective value of the voltage, since the given voltage is the maximal one. For sinusoidal alternating currents (like the standard electrical supply), the effective or 'root mean square' voltage is the maximum voltage divided by the square root of 2 (approximately 1.414). So, the effective voltage (Vrms) is 156 V / √2 = 110.3 V.
Next, we use the formula for power P = V^2 / R, where V is the effective (rms) voltage and R is the resistance. Looking for R, we rearrange the equation to get R = V^2 / P. Substituting the given values we get R = (110.3 V)^2 / 38.8 W = 312.4 Ω. Therefore, the resistance of the light bulb is approximately 312.4 Ω.
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The resistance of a light bulb is approximately 313.57 Ω. This is calculated by converting maximum voltage to RMS voltage and applying the power formula P = V²/R.
To solve this, we need to follow these steps:
First, convert the maximum voltage (Vmax) to the root mean square (RMS) voltage (Vrms) using the formula [tex]V_{rms} = V_{max} / \sqrt2[/tex]. So, [tex]V_{rms} = 156 V / \sqrt2 = 110.3 V[/tex]
Use the power formula that relates power (P), voltage (V), and resistance (R):
[tex]P = V_{rms}^2 / R[/tex]
Rearrange it to solve for resistance (R):
[tex]R = V_{rms}^2 / P[/tex]Substitute the values:
R = (110.3 V)² / 38.8 W
R ≈ 313.57 Ω
Thus, the resistance of the light bulb is approximately 313.57 Ω.
